f
'
(
x
)
=
1
(
x
+
1
)
2
Explanation:
differentiating from first principles
f
'
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
f
'
(
x
)
=
lim
h
→
0
x
+
h
x
+
h
+
1
−
x
x
+
1
h
the aim now is to eliminate h from the denominator
f
'
(
x
)
=
lim
h
=0
(
x
+
h
)
(
x
+
1
)−
x
(
x
+
h
+
1)
h
(
x
+
1
)
(
x
+
h
+
1
)
f
'
(
x
)
=
lim
h
→
0
x
2
+
h
x
+
x
+
h
−
x
2
−
h
x
−
x
h
(
x
+
1
)
(
x+h
+
1
)
f
'
(
x
)
=
lim
h
→
0
h
1
h
1
(
x
+
1
)
(
x
+
h
+1
)
f
'
(
x
)
=
1
(
x
+
1
)
2
Answer:
The surface area of the award is 
Step-by-step explanation:
we know that
The surface area of the square pyramid is equal to the area of the square base plus the area of its four lateral triangular faces
so
![SA=6^{2}+4[\frac{1}{2}(6)(8)]=132\ in^{2}](https://tex.z-dn.net/?f=SA%3D6%5E%7B2%7D%2B4%5B%5Cfrac%7B1%7D%7B2%7D%286%29%288%29%5D%3D132%5C%20in%5E%7B2%7D)
Distribute multiplication over addition and subtraction:
3x+ 12y+10x-5y
Combine like terms:
13 x +7y
15m ^3 - 26m^2 +11m - 4 is the result of the product.
(5m^2 -2m +1)(3m-4)
On opening the bracket, we multiply the terms,
= 15m^3 -6m^2 +3m -20m^2+8m -4
= 15m ^3 - 26m^2 +11m - 4
This is the result of the product
When we mutiply the two equations.
